Rijul saini wrote: Indian Postal Training Set 3, Problem 5 Let $a, b, c$ be integers such that \[\frac ab+\frac bc+\frac ca= 3\]. Prove that $abc$ is a cube of an integer. Use search Problem here

The equation simplifies to $ab^2+bc^2+ca^2=3abc$. We can assume that $\gcd(a,b,c)=1$, otherwise we can divide each of them by their common divisor. Suppose a prime $p\ne3$ divides one of them, say $p|a$. Hence $p|bc^2$, so $p$ divides $b$ or $c$. So any prime divisor $p\ne3$ of $abc$ divides exactly two of $a,b,c$. Suppose WLOG that $p|a,b$. Let $k$ and $l$ be positive integers such that $p^k||a$ and $p^l||b$, also let $a=p^ka_1,b=p^lb_1$. The equation becomes \[p^{k+2l}a_1b_1^2+p^lb_1c^2+p^{2k}ca_1^2=3p^{k+l}a_1b_1c\] Therefore $p^{k+l}||p^lb_1c^2+p^{2k}ca_1^2$. If $l<2k$, then $p^k|b_1c^2+p^{2k-l}ca_1^2$, which is impossible since $p\not|b_1c^2$. Similarly, $l>2k$ is impossible, hence $l=2k$. Thus we conclude that $p^{3k}||abc$. If $3|a$, then $3|bc^2$. So if $3$ divides one of them, it must divide exactly two of $a,b,c$. WLOG suppose $3|a,b$, let $3^x||a,3^y||b$. Let $a=3^xa_2,b=3^yb_2$. The equation becomes \[3^{x+2y}a_2b_2^2+3^yb_2c^2+p^{2x}ca_2^2=3^{x+y+1}a_2b_2c\] Since $y\ge1$, then $x+2y\ge x+y+1$. Thus $3^{x+y+1}||3^yb_2c^2+p^{2x}ca_2^2$. Similarly as above, we can prove that $y=2x$. Therefore $3^{3x}||abc$. Thus, for all prime divisors $p$ of $abc$, we have $p^{3k}||abc$ for some positive integer $k$. This implies that $abc$ is a perfect cube.

I am rephrasing the problem here. Prove that if $a,b,c$ are positive integers and $gcd(a,b,c) = 1$, then $a=b=c=1$ is the only solution to $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 3$.

nicetry007 wrote: I am rephrasing the problem here. Prove that if $a,b,c$ are positive integers and $gcd(a,b,c) = 1$, then $a=b=c=1$ is the only solution to $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 3$. Your problem is true, but it's not a rephrasing of the original problem as $a,b,c$ are not necessarily positive.

Let wlog $gcd(a,b,c)=1$. Then $a|bc$ from the equation,$a,b,c=\pm 1$ and checking shows that $a=b=c=1$

Sorry I am wrong. But how did nicetry conclude $a=b=c=1$ in positive integers

Using AM GM its easy to show that $ a=b=c $

YESMAths wrote: Using AM GM its easy to show that $ a=b=c $ AM-GM only works for nonnegatives...

Oh sorry but still it works for the nonnegative part and thanks I didnt read the question carefully for being defined as integers

jgnr wrote: The equation simplifies to $ab^2+bc^2+ca^2=3abc$. We can assume that $\gcd(a,b,c)=1$, otherwise we can divide each of them by their common divisor. Suppose a prime $p\ne3$ divides one of them, say $p|a$. Hence $p|bc^2$, so $p$ divides $b$ or $c$. So any prime divisor $p\ne3$ of $abc$ divides exactly two of $a,b,c$. Suppose WLOG that $p|a,b$. Let $k$ and $l$ be positive integers such that $p^k||a$ and $p^l||b$, also let $a=p^ka_1,b=p^lb_1$. The equation becomes \[p^{k+2l}a_1b_1^2+p^lb_1c^2+p^{2k}ca_1^2=3p^{k+l}a_1b_1c\]Therefore $p^{k+l}||p^lb_1c^2+p^{2k}ca_1^2$. If $l<2k$, then $p^k|b_1c^2+p^{2k-l}ca_1^2$, which is impossible since $p\not|b_1c^2$. Similarly, $l>2k$ is impossible, hence $l=2k$. Thus we conclude that $p^{3k}||abc$. If $3|a$, then $3|bc^2$. So if $3$ divides one of them, it must divide exactly two of $a,b,c$. WLOG suppose $3|a,b$, let $3^x||a,3^y||b$. Let $a=3^xa_2,b=3^yb_2$. The equation becomes \[3^{x+2y}a_2b_2^2+3^yb_2c^2+p^{2x}ca_2^2=3^{x+y+1}a_2b_2c\]Since $y\ge1$, then $x+2y\ge x+y+1$. Thus $3^{x+y+1}||3^yb_2c^2+p^{2x}ca_2^2$. Similarly as above, we can prove that $y=2x$. Therefore $3^{3x}||abc$. Thus, for all prime divisors $p$ of $abc$, we have $p^{3k}||abc$ for some positive integer $k$. This implies that $abc$ is a perfect cube. Wrong solution please ignore

mathmdmb wrote: Sorry I am wrong. But how did nicetry conclude $a=b=c=1$ in positive integers we let d=gcd(a,b,c) we have the equation ab^2 + ca^2 + bc^2 =3abc so if we let a=dx b=dy and c=dz and rewrite the equation we get d^3xy^2 + d^3zx^2 + d^3yz^2 =d^3*3xyz cancelling out d^3 we have, xy^2 +zx^2 +yz^2 = 3xyz and here gcd(x,y,z)=1. so we just need to solve this equation and we have x|yz , y|xz , z|xy as we have x,y,z as integers so (x,y,z)=(1,1,1) or (x,y,z)=(-1,-1,-1) and so a*b*c is always a cube. Hence Proved.